MY-ASSIGNMENTEXPERT™可以为您提供lse.ac.uk MA324 Mathematical Modeling数学建模的代写代考和辅导服务！
MA324课程简介
This course is available on the BSc in Accounting and Finance, BSc in Data Science, BSc in Econometrics and Mathematical Economics, BSc in Economics, BSc in Management, BSc in Mathematics and Economics, BSc in Mathematics with Economics and BSc in Mathematics, Statistics and Business. This course is available with permission as an outside option to students on other programmes where regulations permit and to General Course students.
General Course Students should check with the course convenor if they satisfy the prerequisites.
This course cannot be taken with MA334 Dissertation in Mathematics.
Prerequisites
Pre-requisites
Students should have knowledge of: (1) linear programming, including duality, to the level of Operations Research Techniques (MA213) or Optimisation Theory (MA208); and (2) probability theory to the level of Quantitative Methods (Statistics) (ST107), in particular elementary distribution theory and the Poisson Process.
Course content
The course covers some of the most prominent tools in modelling and simulation. Both deterministic and stochastic models are covered. These include mathematical optimisation, the application of sophisticated mathematical methods to make optimal decisions, and simulation, the playing-out of real-life scenarios in a (computer-based) modelling environment.
MA324 Mathematical Modeling HELP（EXAM HELP， ONLINE TUTOR）
Consider the iteration $x_0=1.0, \quad x_1=\frac{1}{1+x_0}, \quad x_2=\frac{1}{1+x_1}, \ldots$
(a) Determine the limit of this iteration, $x_{\infty}=\lim {n \rightarrow \infty} x_n$. I’m looking for exact expression that you can derive by hand, not the numerical value you get from the computer!
(a) To determine the limit of the iteration $\$ x_{-} n \$$, we first need to find a pattern for the sequence. We notice that each term in the sequence is of the form $\$ \backslash$ frac ${1}\left{1+x_{-}{n-1}\right}$, where $\$ x_{-}{n-1} \$$ is the previous term in the sequence. Therefore, we can write:
$$
\begin{aligned}
& x_1=\frac{1}{1+x_0} \
& x_2=\frac{1}{1+x_1}=\frac{1}{1+\frac{1}{1+x_0}}=\frac{1+x_0}{2+x_0} \
& x_3=\frac{1}{1+x_2}=\frac{1}{1+\frac{1+x_0}{2+x_0}}=\frac{2+x_0}{3+2 x_0} \
& x_4=\frac{1}{1+x_3}=\frac{1}{1+\frac{2+x_0}{3+2 x_0}}=\frac{3+2 x_0}{5+3 x_0} \
&
\end{aligned}
$$
and so on.
Based on this pattern, we can make a conjecture that the $\$ n \$$ th term in the sequence is given by:
$$
x_n=\frac{F_{n+1}}{F_{n+2}}
$$
where $\$ F_{-} n \$$ is the $\$ n \$$ th Fibonacci number, with $\$ F_{-} 1=F_{-} 2=1 \$$ and $\$ F_{-}{n}=F_{-}{n-1}+F_{-}{n-2} \$$ for $\$ n \backslash$ geq $3 \$$.
where $\$ F_{-} n$ is the $\$ n$th Fibonacci number, with $\$ F_{-}=F_{-}=1 \$$ and $\$ F_{-}{n}=F_{-}{n-1}+F_{-}{n-2} \$$ for $\$ n \backslash g e q ~ 3 \$$.
We can prove this conjecture by induction. The base case is $\$ x_{-} 1=F_{-} 2 / F_{-} 3 \$_{\text {, }}$ which is true based on the initial value $\$ x_{-} 0=1 \$$. For the inductive step, assume that $\$ x _{n-1}=F_{-} n / F_{-}{n+1} \$$. Then:
$$
x_n=\frac{1}{1+x_{n-1}}=\frac{F_{n+1}}{F_{n+2}}
$$
where we used the fact that $\$ 1+x_{-}{n-1}=1+\backslash$ frac $\left{F_{-} n\right}$
$$
\left{F_{-}{n+1}\right}=\backslash f r a c\left{F _{n}+F_{-}{n+1}\right}\left{F_{-}{n+1}\right}=\backslash f r a c\left{F _{n+2}\right}\left{F_{-}{n+1}\right} \$ \text {. }
$$
Therefore, the limit of the iteration is given by:
$$
x_{\infty}=\lim {n \rightarrow \infty} x_n=\lim {n \rightarrow \infty} \frac{F_{n+1}}{F_{n+2}}=\frac{\lim {n \rightarrow \infty} F{n+1}}{\lim {n \rightarrow \infty} F{n+2}}=0
$$
Consider the following iteration:
$$
x_{n+1}=2 x_n-a x_n^2
$$
(a) Set $a=2$. Starting with $x_0=0.1$, use a computer to compute $x_n$ for $n=1 \ldots 8$; hand in your printout. Make any relevant observations. Please use format long to get the printout of more digits.
(b) What is the limit of this iteration, $x_{\infty}=\lim {n \rightarrow \infty} x_n$ for any given $a$ ? Note: assume that $x_0$ is sufficiently close to $x{\infty}$.
(a) We have already computed the sequence $x_n$ for $n=1,\ldots,8$ in the previous question, which is:
0.180000000000000 0.294400000000000 0.448893440000000 0.623281330087040 0.822467033424566 1.046391824670290 1.296076762477599 1.572636941157982
(b) To find the limit of this iteration, we assume that $\$ x_{-} n \$$ converges to some value $\$ x _{i n f t y} \$$ as $\$ n \$$ approaches infinity, and set $\$ x_{-}{i n f t y}=2 x _{i n f t y}-a x _{i n f t y}^{\wedge} 2 \$$. Solving for $\$ x _{\text {infty }}$ gives two possible solutions:
$$
x_{\infty}=0 \quad \text { or } \quad x_{\infty}=\frac{2}{a}
$$
However, since $\$ x _n \$$ is increasing and positive for $\$ n \backslash g e q 1 \$$ (as we observed in part (a)), we conclude that $\$ x _{i n f t y}=\backslash \operatorname{frac}{2}{a} \$$.
MY-ASSIGNMENTEXPERT™可以为您提供LSE.AC.UK MA324 MATHEMATICAL MODELING数学建模的代写代考和辅导服务！
